A particle is in a harmonic oscillator potential. Which are the possible spreads for a simultaneous measurement of the momentum? 
A particle is in a harmonic oscillator potential, not in the ground state. The position of the particle is known with an rms spread of $1\mathring{\text{A}}$. Which of the following are possible spreads for a simultaneous measurement of the momentum?
  
  
*
  
*$2.63 \times 10^{−25} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  
*$5.27 \times 10^{−25} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  
*$1.05 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  
*$2.10 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  


I think the answer is $5.27 \times 10^{−25} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$ from which I used the minimum uncertainty relation $$\Delta x \Delta p= \frac{\hbar}{2}$$ With $\Delta x=10^{-10}$ so $$\Delta p=\frac{\hbar}{2\Delta x}\approx\frac{1.055\times 10 ^{-34}}{2\times 10^{-10}}\approx 5.275\times 10^{-25}\mathrm{\,kg\,m}\, \mathrm{s}^{−1}$$

The correct answers are 


  
*$1.05 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  
*$2.10 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  


I note that my (wrong) answer is precisely double another answer on that list $$\frac{5.275\times 10^{-25}}{2}=2.6375\mathrm{\,kg\,m}\, \mathrm{s}^{−1}$$

I am very curious as to why there must be two values that are connected somehow. I managed to use the uncertainty relation to get one answer, but how on Earth do you get the other answer? 
More importantly for now, could anyone please explain or give hints so that I can understand why the correct answers are 

$$1.05 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1} \qquad \text{&} \qquad 2.10 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}\,?$$

 A: It says it's not in the ground state. You can only get the lowest bound in the Heisenberg Uncertainty Principal when the harmonic oscillator is in the ground state. Since we are not in the ground state, we know the spread must be larger than the number you calculate.
In other words, since it says the system is not in the ground state, you cannot assume minimum uncertainty. The correct answers are therefore all answers larger than $5.275 \times 10^{-25} kg\cdot m \cdot s^{-1}$ (Answers 3&4).
There is not any more information to actually calculate answers 3&4. Any choice the question would have given larger than the minimum spread would be correct. There are no calculations to be done to get those two choices. You did everything right in the math, just not in the interpretation of the math given the assumptions stated in the problem.
There are an infinite amount of possible correct answers. You just have to pick the two that the maker of the question gave that are larger than the spread from the minimum uncertainty.
